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    Show that the equation x(x-2)^2=3 has only one real root and state the value of this root.

    an idea is kind of forming in my head but I'm still quite stuck, how would I begin to tackle this?
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    (Original post by Martynnnn)
    Show that the equation x(x-2)^2=3 has only one real root and state the value of this root.

    an idea is kind of forming in my head but I'm still quite stuck, how would I begin to tackle this?
    One way of doing it:
    Draw the graphs of (x-2)^2 and 3/x to show there is only one point.
    You will see they intersect at one point and one point only.
    The word state suggests you should be able to so see the solution from the graph and 'deduce' its 3.
    You can of course use the factor theorem.
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    Expand it, put everything on one side, find a factor. Show that the remainder when you divide the equation by the factor doesn't have any real roots by finding the discriminant.
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    (Original post by Damask-)
    Expand it, put everything on one side, find a factor. Show that the remainder when you divide the equation by the factor doesn't have any real roots by finding the discriminant.
    It's C1. Factor theorem is not (to my knowledge on any exam board) introduced yet hence the suggestion of a sketch and deducing the value.
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    (Original post by Martynnnn)
    Show that the equation x(x-2)^2=3 has only one real root and state the value of this root.

    an idea is kind of forming in my head but I'm still quite stuck, how would I begin to tackle this?
    Take it all to one side, then use b^2-4ac which should equal 0.
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    (Original post by m4ths/maths247)
    It's C1. Factor theorem is not (to my knowledge on any exam board) introduced yet hence the suggestion of a sketch and deducing the value.
    WJEC have remainder theorem (and thus factor theorem) on C1. Although I don't think this is WJEC anyway.
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    draw the graph y = x(x-2)(x-2)...find the local maximum by differentiating... show that the line y = 3 passes above the local maximum...
 
 
 
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